# [Haskell-cafe] lawless instances of Functor

Derek Elkins derek.a.elkins at gmail.com
Mon Jan 4 18:01:46 EST 2010

```On Tue, Jan 5, 2010 at 7:14 AM, Paul Brauner <paul.brauner at loria.fr> wrote:
> Hi,
>
> I'm trying to get a deep feeling of Functors (and then pointed Functors,
> Applicative Functors, etc.). To this end, I try to find lawless
> instances of Functor that satisfy one law but not the other.
>
> I've found one instance that satisfies fmap (f.g) = fmap f . fmap g
> but not fmap id = id:
>
> data Foo a = A | B
>
> instance Functor Foo where
>  fmap f A = B
>  fmap f B = B
>
> -- violates law 1
> fmap id A = B
>
> -- respects law 2
> fmap (f . g) A = (fmap f . fmap g) A = B
> fmap (f . g) B = (fmap f . fmap g) B = B
>
> But I can't come up with an example that satifies law 1 and not law 2.
> I'm beginning to think this isn't possible but I didn't read anything
> saying so, neither do I manage to prove it.
>
> I'm sure someone knows :)

Ignoring bottoms the free theorem for fmap can be written:

If h . p = q . g then fmap h . fmap p = fmap q . fmap g
Setting p = id gives
h . id = h = q . g && fmap h . fmap id = fmap q . fmap g
Using fmap id = id and h = q . g we get,
fmap h . fmap id = fmap h . id = fmap h = fmap (q . g) = fmap q . fmap g

So without doing funky stuff involving bottoms and/or seq, I believe
that fmap id = id implies the other functor law (in this case, not in
the case of the general categorical notion of functor.)
```